Q.6.111 - Elementary Statistics

Question

Booted Eagles. The rare booted eagle of western Europe was the focus of a study by S. Suarez et al. to identify the optimal nesting habitat for this raptor. According to their paper "Nesting Habitat Selection by Booted Eagles (Hieraaefus pennatus) and Implications for Management" (Journal of Applied Ecology, Vol. 37. pp. 215-223). the distances of such nests to the nearest marshland are normally distributed with a mean $4.66 \mathrm{km}$ and a standard deviation $0.75 \mathrm{km}$. Let $Y$ be the distance of a randomly selected nest to the nearest marshland.

Determine and interpret

a. $P(Y>5)$.

b. $P(3\le Y\le 6)$.

Step-by-Step Solution

Answer

a). The probability that a randomly selected nest is more than $5$ kilometres away from the nearest marshland is $0.3264$.

b). The probability that the distance of a randomly selected nest to the nearest marshland is between $3 \mathrm{km}$ and $6 \mathrm{km}$ is $0.9497$.

Part (a) Step 1: Given Information

The distances of such nests to the nearest marshland are normally distributed with mean $4.66 \mathrm{km}$ and standard deviation $0.75 \mathrm{km}$. Let $Y$ be the distance of a randomly selected nest to the nearest marshland.

Part (a) Step 2: Explanation

we have $\mu =4.66 \mathrm{km}$ and $\sigma =0.75 \mathrm{km}$

$P(Y>5)$ Indicate the probability that a randomly chosen nest is more than $5$ miles away from the nearest marshland.

The figure shows the required shaded region.

Part (a) Step 3: Explanation

We need to compute the z-score for the $y$-value $5$ :

$y =5 \to$

$z=\frac{5-\mu }{\sigma }$

$=\frac{5-4.66}{0.75}$

$=0.45$

We have to determine the standard normal curve that lies below $0.45$. The area to the left of $0.45$ it $0.6736$.The area to the right of $0.45$ is $1-0.6736=0.3264$. The required area, shaded in the figure, is $0.3264$

Part (b) Step 4: Given Information

The distances of such nests to the nearest marshland are normally distributed with mean $4.66 \mathrm{km}$ and standard deviation $0.75 \mathrm{km}$. Let $Y$ be the distance of a randomly selected nest to the nearest marshland.

Part (b) Step 5: Explanation

The normal curve for the variable is depicted in the following diagram. The tick marks are separated by units, which means that the distance between each tick mark is equal to the standard deviation.

The required shaded zone and its delimiting $y$-values, $3$ and $6$, are shown in the diagram.

Part (b) Step 3: Explanation

We need to compute the $z$-scores for the $y$-values $3$ and $6$ :

$y =3 \to$

$z=\frac{3-\mu }{\sigma }$

$=\frac{3-4.66}{0.75}$

$=-2.21$

And

$y=6\to$

$=\frac{6-\mu }{\sigma }$

$=\frac{6-4.66}{0.75}$

$=1.79$

The z- scores are marked beneath the $y$-values in the figure.

We need to find the area between the two points on the standard normal curve $-2.21$ and $1.79$. The area to the left of $-2.21$ is $0.0136$ and the area to the left of $1.79$ it $0.9633$. The required area, shaded in the figure, is therefore $0.9633-0.0136=0.9497$.